MCQ Questions for Class 12 Medical Physics Atoms Quiz 7

The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about

$10^{-40}$ . An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

0%

The radius of the first Bohr orbit and the estimated size of the whole universe is same.
0%

The radius of the first Bohr orbit is much greater than the estimated size of the whole universe.
0%

The radius of the first Bohr orbit is much smaller than the estimated size of the whole universe.
0%
None of the above.

Explanation

Radius of the first Bohr orbit is given by the relation,

$r_1=\dfrac{4\pi \epsilon_0(\dfrac{h}{2\pi})^2}{m_e e^2}$ ......(1)

Where,

$\epsilon_0=$ Permittivity of free space

$h=$ Planck's constant

$m_e=$ Mass of an electron

$e=$ Charge of an electron

$m_p=$ Mass of a proton

$r=$ Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as:

$F_C=\dfrac{e^2}{4\pi \epsilon_0 r^2}$ .....(2)

Gravitational force of attraction between an electron and a proton is given as:

$F_G=\dfrac{Gm_p m_c}{r^2}$ .....(3)

Where

$G=$ Gravitation constant

$=6.67 \times 10^{-11}\,Nm^2/kg^2$

If an electrostatic (Coulomb) force and the gravitation force between an electron and a proton are equal, then we can write:

$F_G=F_C$

$\dfrac{Gm_p m_c}{r^2}=\dfrac{e^2}{4\pi \epsilon_0 r^2}$

$\therefore \dfrac{e^2}{4\pi \epsilon_0}=Gm_p m_c$ .....(4)

Putting the value of equation ( 4 ) in equation ( 1 ), we get:

$r_1=\dfrac{(\dfrac{h}{2\pi})^2}{Gm_pm_c^2}$

$=\dfrac{(\dfrac{6.63\times 10^{-34}}{2\times 3.14})^2}{6.67\times 10^{-11}\times 1.67 \times 10^{-27}\times (9.1\times 10^{-31})^2}\approx 1.21\times 10^{29}\,m$

It is known that the universe is

$156$ billion light years wide or

$1.5\times 10^{27}\, m$ wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Who gave the Quantum model of hydrogen atom ?

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S.N. Bose
0%
Niels Bohr
0%
James clerk Maxwell
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R. A. Millikan

Rutherford's alpha-particle scattering experiment was responsible for the discovery of:

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atomic nucleus
0%
electron
0%
proton
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neutron

Which of the following is not an inherent property of the photons?

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Momentum
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Energy
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Charge
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Velocity

Explanation

A photon is a particle which has zero charge and zero mass and is denoted by

$\gamma$ . The energy of photon is

$E = hv$
Here,

$v =$ frequency and

$h =$ Planck's constant.
The momentum of the photon is

${h}/{v}$ and its velocity is the velocity of light

$\left(c\right)$ .
So, the charge is not the property of photons.

The angular speed

$(\omega)$ of an electron revolving in

$n^{th}$ Bohr orbit and corresponding principal quantum number(n) are related as ____________.

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$\omega\propto n^3$
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$\omega\propto\dfrac{1}{ n^3}$
0%
$\omega\propto n^2$
0%
$\omega\propto\dfrac{1}{ n^2}$

Explanation

Bohr's Model gives the following conclusions,

$v\propto\dfrac{1}{n}$

$\Rightarrow r\propto{n^2}$

$\Rightarrow \omega \propto \dfrac vr \propto \dfrac {1}{n^3}$

The ionisation potential of hydrogen atom is -13.6 eV. An electron in the ground state of a hydrogen atoms absorb a photon of energy 12.75 eV. How many different spectral line can one expect when the electron make a downward transition?

0%
1
0%
4
0%
2
0%
6

Explanation

$\displaystyle E={ { E }_{ 1 } }/{ { n }^{ 2 } }$
Energy used for excitation is 12.75 eV
ie, (-13.6+12.75)eV = -0.85eV image
Energy levels of H-atom
The photons of energy 12.75eV can excite the fourth level of H-atom. Therefore, six lines will be emitted.

$\left( n\frac { \left( n-1 \right) }{ 2 } lines \right)$ .

If a cathode ray tube has a potential difference

$V$ volt between the cathode and anode, then the speed

$v$ of cathode rays is given by:

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$v\propto { V }^{ 2 }$
0%
$v\propto \sqrt { V }$
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$v\propto \cfrac { 1 }{ V }$
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$v\propto V$

Explanation

$qV=\cfrac { 1 }{ 2 } m{ v }^{ 2 }$
or

$v=\sqrt { \cfrac { 2qV }{ m } }$
ie

$v\propto \sqrt { V }$

If an electron is revolving around the hydrogen nucleus at a distance of

$0.1$ nm, what should be its speed?

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$2.188\times 10^6 ms^{-1}$
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$1.094\times 10^6 ms^{-1}$
0%
$4.376\times 10^6 ms^{-1}$
0%
$1.59\times 10^6 ms^{-1}$

Explanation

electrostatic force=centripetal force

$\displaystyle\frac{1}{4\pi \varepsilon_0}\frac{Ze^2}{r^2}=\frac{mv^2}{r}$

$\therefore v=\sqrt{\left(\displaystyle\frac{1}{4\pi \varepsilon_0}\frac{Ze^2}{mr}\right)}$

$=\sqrt{\displaystyle\frac{9\times 10^9\times 1\times (1.6\times 10^{-19})^2}{(9.1\times 10^{-31})\times (0.1\times 10^{-9})}} =1.59\times 10^6ms^{-1}$

In a hydrogen atom, the electron is making

$6.6\times 10^{15} revs^{-1}$ around the nucleus in an orbit of radius

$0.528 \overset{o}{A}$ . The magnetic moment

$(A-m^2)$ will be.

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$1\times 10^{-15}$
0%
$1\times 10^{-10}$
0%
$1\times 10^{-23}$
0%
$1\times 10^{-27}$

Explanation

Current,

$I=6.6\times 10^{15}\times 1.6\times 10^{-19}$

$=10.5\times 10^{-4}A$ .
Area

$A=\pi R^2=3.142\times (0.528)^2\times 10^{-20}m^2$
so, Magnetic moment

$M=IA=10.5\times 10^{-4}\times 3.142\times (0.528)^2\times 10^{-2}$

$=10\times 10^{-24}=10^{-23}$ units

The spectrum obtained from the chromosphere of the sun at the time of total solar eclipse is :

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line emission spectrum
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band emission spectrum
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continuous emission spectrum
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line absorption spectrum

Explanation

The spectrum obtained is the line emission spectrum of

${ D }_{ 3 }$ line of neutral helium at

$5876 \mathring{A}$ wavelength (almost yellow colour).

Which state of triply ionised Beryllium

$(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen ?

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$n = 3$
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$n = 4$
0%
$n = 1$
0%
$n = 2$

Explanation

From Bohr's atomic model, atomic radius for

$n$ th state is given by the formula,

$r=0.059\dfrac { { n }^{ 2 } }{ z } A$

Since for Be,

$z=4$

$\dfrac { { n }^{ 2 } }{ 4 } =\dfrac { { 1 }^{ 2 } }{ 1 }$

Hence,

$n=2$

Bohr's atom model assumes.

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The nucleus is of infinite mass and is at rest
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Electrons in a quantized orbit will not radiate energy
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Mass of electron remains constant
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All the above.

The difference between the wavelengths of the Stokes line and Anti-Stokes lines in the Raman spectrum of

$H-Br$ molecule is

$100\mathring { A }$ . If the wavelength of the Anti-Stokes line is

$5000\mathring { A }$ the wavelength of the incident radiation is:

0%
$5050 \mathring { A }$
0%
$4950 \mathring { A }$
0%
$5100 \mathring { A }$
0%
$4900 \mathring { A }$

Explanation

Stokes and antistokes lines are symmetrically placed about main line.

Therefore if the difference between the wavelengths of a stokes line and its corresponding antistokes lines is

$100 \mathring { A }$ , then the difference between antistokes line and main line is

$50 \mathring { A }$ .

More over wavelength of antistokes line is less than main line.

$\therefore$ wavelength of incident radiation is

$5050 \ \mathring { A }$ .

Most of the

$\alpha$ -particles passed straight through the gold foil. This proved the existence of ________ in an atom.

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electrons
0%
nucleus
0%
empty space
0%
positive charge

Explanation

In Rutherford's gold foil experiment for the discovery of atomic structure, most of the

$\alpha$ -particles passed straight (around 99% non-deflected

$\alpha-$ particles) through the gold foil which proved the existence of empty space in an atom.

If elements with principal quantum number

$n>4$ were not allowed in nature, the number of possible elements would have been:

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$32$
0%
$60$
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$64$
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$4$

Explanation

Hint:

Add all the elements from

$n=1$ to

$n=4$ .

Step 1: Write the number of elements for each shell.

The number of elements for

$n=1$ are

$2$ .

The number of elements for

$n=2$ are

$8$ .

The number of elements for

$n=3$ are

$18$ .

The number of elements for

$n=4$ are

$32$ .

Step 2: adding all the elements,

from

$n=1$ to

$n=4$ .

Total number

$= 2+8+18+32=60$

$\textbf{Hence option B correct}$

Which of the following is not a conclusion of Rutherford experiment?

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An atom has a very small compact nucleus
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An atom is mainly empty space
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An atom's mass is concentrated in the nucleus
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An atom has a very dense nucleus
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An atom has a negatively charged nucleus

Explanation

The statement

$(E)$ is not a conclusion of Rutherford experiment
An atom has a positively charged nucleus.
The following are the conclusions from Rutherford’s scattering experiment.
1. Many alpha particles pass without deflection. Atom has most of the space as empty.
2. Some alpha- particles deflect by

$90^{o}$ or even larger angles. This is due to repulsion in like charges. Hence, the heavy core of the atom is positively charged and is termed the nucleus.
3. Very few alpha particles bounced back. The nucleus occupies a very small volume as compared to atomic volume.
4. The alpha- particles having significant mass are deflected by the nucleus. Hence almost the entire mass of the atom is concentrated at its center which is the nucleus.

Hence. option

$E$ is the correct choice here.

For the Paschen series the values of

${ n }_{ 1 }$ and

${ n }_{ 2 }$ in the expression

$\Delta E=Rhc\left( \dfrac { 1 }{ { n }_{ 1 }^{ 2 } } -\dfrac { 1 }{ { n }_{ 2 }^{ 2 } } \right)$ are

0%
${ n }_{ 1 }=1,{ n }_{ 2 }=2,3,4,\dots$
0%
${ n }_{ 1 }=2,{ n }_{ 2 }=3,4,5,\dots$
0%
${ n }_{ 1 }=3,{ n }_{ 2 }=4,5,6,\dots$
0%
${ n }_{ 1 }=4,{ n }_{ 2 }=5,6,7,\dots$

Explanation

Paschen series is the third series of hydrogen spectrum and lies in the infrared region, therefore for this series,

${ n }_{ 1 }=3$ and

${ n }_{ 2 }>{ n }_{ 1 }$ , ie,

$4,5,6,\dots$

Which of the following was not a conclusion of Rutherford's gold foil experiment?

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The atom is mainly empty space.
0%
The nucleus has a negative charge.
0%
The atom has a dense nucleus.
0%
Alpha particles can pass through a thin sheet of gold foil.
0%
All of the above are correct regarding the gold foil experiment.

Explanation

The Rutherford gold leaf experiment concluded that most (99%) of all the mass of an atom is in the nucleus of the atom, that the nucleus is very small ( $10^5$ times small than the size of the atom) and that is is positively charged.

Statement I: Rutherford in his experiment observed that, most of the alpha particles passed straight through a thin sheet of gold foil.
Statement II: Most of the space inside the atom is empty.

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Both statements I and II are true.
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Statement I is true and statement II is false.
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Statement I is false and statement II is true.
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Both statement I and II are false.
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Ambiguous

Explanation

Most of the space inside the atom is empty. This proved to be true when Rutherford observed that most of the

$\alpha$ -particles passed through the gold foil without getting deflected.
The above statements are true and one of the conclusions of Rutherford's experiment.

For absorption and emission spectra of atoms which one of the following particle changes energy levels?

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Alpha
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Electron
0%
Neutron
0%
Photon
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Proton

$_{ 95 }^{ 241 }{ Am }\ \rightarrow \ _{ 93 }^{ 237 }{ Np }+Y$
In the above nuclear fusion reaction, Identify the particle represented by

$Y$ ?

0%
A proton
0%
An electron
0%
An alpha particle
0%
A gamma ray
0%
A beta particle

Explanation

The given nuclear reaction :

$^{241}_{95} Am \rightarrow ^{237}_{93} Np$

$+$

$^A_Z Y$

In a nuclear reaction, the mass number as well as the atomic number is conserved.

$\therefore$

$241 = 237 + A$

$\implies A =4$

Also,

$95 = 93 + Z$

$\implies Z =2$

Thus, the particle

$Y$ is an alpha particle i.e

$^4_2He$ .

Which of the following statement is true on a scattering of atom according to Rutherfords experiments on the alpha particle ?

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Atoms are roughly spherical with a radius of about ${10}^{-10} m$
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The electrons occupy quantized energy levels, absorbing or emitting energy only when they make a quantum jump between these levels
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The density of positive charge within an atom is not uniform throughout the atoms volume
0%
Allowed electron orbits must have a circumference equal to a whole number times the electrons de Broglie wavelength
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Alpha particles are positively charged

Statement 1: A large number of alpha particles were deflected in the Rutherford experiment.
Statement 2: Alpha particles that came close to the nucleus were deflected.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
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Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
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Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 is not correct.

Explanation

Statement 1 not correct because only a few alpha particles were deflected in the Rutherford experiment. Most alpha particles passed without deflection as most part of the atom is empty.
Statement 2 is correct because alpha particles that came close to the positively charged nucleus were deflected.

Hence, option

$C$ is correct i.e. Statement 1 is not correct but Statement 2 is correct.

Which of the following is true regarding atoms according to Rutherford's gold foil experiment?

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Are mostly space
0%
Are in continuous motion
0%
Have negative orbitals
0%
Have diffuse charge distribution
0%
Have dense crystalline structure

Explanation

In Rutherford's experiment most of the

$\alpha$ -particles were found passing the foil without any deflection which means atoms have a lot of empty space.

According to Bohr's model of the atom, atoms emit or absorb radiation only at certain wavelengths.Identify why is it so

0%
Because the protons and electrons are distributed evenly throughout the atom
0%
Because electrons can orbit the nucleus at any radius
0%
Because electrons orbit the nucleus only at certain discrete radii
0%
Because protons orbit the nucleus only at certain discrete radii
0%
Because photons can only have discrete wavelength

Explanation

Bohr's frequency condition is given by

$hc/\lambda=E_{2}-E_{1}$

Above relation shows that wavelength

$\lambda$ of radiation depends upon energy difference of two orbits,

$\lambda$ is certain because energy difference is constant for two orbits it means

$E_{1}$ and

$E_{2}$ are also fixed i.e. electron is revolving in some fixed orbits of discrete radii. These orbits are called stationary orbits in which electron does not emit energy when revolving in these orbits.

The discrete spectral lines of line spectra occur when excitation of electrons takes place in

0%
Solids
0%
Liquids
0%
Gases
0%
Plastics
0%
All of these

Which of the following assumptions made by Neils Bohr regarding the hydrogen atom was NOT new and "radical"?

0%
Electrons are only stable in certain orbits
0%
Atoms only emit electromagnetic radiation when electrons transition to lower energy states
0%
The Coulomb force is responsible for keeping the electron in orbit around the positive nucleus
0%
An electron in a hydrogen atom can only orbit the nucleus if its angular momentum is some integer multiple of $h/2 \pi$
0%
All of the above were new and "radical"

When a high energy (fast moving) electron approaches a heavy metal target, some of the electrons are significantly deflected by nuclei, and they lose much of their energy. This lost energy goes into the creation of high energy photons.
What process is being described here?

0%
Compton scattering
0%
The photoelectric effect
0%
X-ray production
0%
Pair production
0%
Quantum tunneling

A high energy photon collides elastically with a charged particle (usually an electron)., The photon "bounces away" with less energy (a longer wavelength).
The process described here is known as :

0%
Compton scattering
0%
The photoelectric effect
0%
X-ray production
0%
Annihilation
0%
Quantum tunneling

Rutherford's theory assumed which of the following paths for the electron?

0%
Elliptical
0%
Circular
0%
Spiral
0%
All the above

Rutherford's experiment on scattering of

$\alpha$ -particles showed for the first time that the atom has

0%
Nucleus
0%
Electron
0%
Proton
0%
Neutron

A beam of alpha-particles is fired at a piece of gold foil as shown in the image. After striking the gold foil in which direction do most of the alpha-particles travel?

0%
$P$
0%
$Q$
0%
$R$
0%
$S$

Explanation

Alpha particles are positively charged particles made up of two protons, two neutrons, and zero electrons. Due to the fact that protons have a +1 charge and neutrons hold no charge. This would give the particle a +2 charge overall.
Originally, Rutherford thought that particles would fly straight through the foil. However, he found that the particles would be shifted or deflected when passing through the foil. This is due to the fact that like charges repel each other.
As the positively charged alpha particle would fly through the foil, it would come in proximity with a positively charged nucleus of an atom.
Thus we can conclude that the particle travels in

$P$ direction.

Ernest Rutherford performed an experiment to know the arrangement of electrons within an atom. In this experiment, fast-moving alpha particles were made to fall on a thin gold foil(100 nm thickness). The

$\alpha$ -particles are doubly charged helium ions. Since they have a mass of 4 u, the fast-moving

$\alpha$ -particles have a considerable amount of energy. It was expected that

$\alpha$ -particles would be deflected by subatomic particles in the gold atom. Since

$\alpha$ -particles were much heavier than protons, he did not expect to see large deflections.

But, the

$\alpha$ -particle scattering experiment gave totally unexpected results.
The following observations were made:
(i) Most of the fast-moving -particles passed straight through the gold foil.
(ii) Some of the -particles were deflected by the foil by small angles.
(iii) Surprisingly one out of every 12000 particles appeared to rebound.

Following are the observations which conclude the presence of the nucleus at the center of the atom:

1) Some of the alpha rays are deflected back with some angle. (presence of nucleus)

2) Very few alpha particles are bounced back with zero-degree angles. (Alpha rays hit the center of the nucleus)

Hence option

A,C,DA,C,D are correct.

Which of the following observations of

$\alpha$ -ray scattering experiment lead to the presence of small positively charged nucleus in the centre?

0%
Most of the $\alpha$ -particles passed straight
0%
Most of the $\alpha$ -particles rebounded after hitting the atoms
0%
Only a few $\alpha$ -particles deflected away from their path
0%
Very few $\alpha$ -particles rebounded

Explanation

Following are the observations which conclude the presence of the nucleus at the center of the atom:

1) Some of the alpha rays are deflected back with some angle. (presence of nucleus)

2) Very few alpha particles are bounced back with zero-degree angles.(Alpha rays hit the center of the nucleus)

Hence option

$A,C ,D$ are correct.

Rutherford's experiment for the first time proved that an atom has:

0%
protons
0%
a nucleus
0%
neutrons
0%
electrons

As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom

0%
decreases
0%
increases
0%
remains the same
0%
does not increase.

Explanation

Total energy of the electron in nth orbit

$E_n = \dfrac{-13.6}{n^2} \ eV$
Potential energy

$P.E = 2E_n = \dfrac{-27.2}{n^2} \ eV$
Thus potential energy of the atom increases with increase in

$n$ .

In which of the following transitions will the wavelength be minimum ?

0%
$n = 5$ to $n = 4$
0%
$n = 4$ to $n = 3$
0%
$n = 3$ to $n = 2$
0%
$n = 2$ to $n = 1.$

Explanation

Wavelength of emitted photon due to transition :

$\dfrac{1}{\lambda} = R\bigg(\dfrac{1}{n_2^2}-\dfrac{1}{n_1^2}\bigg)$
(A) :

$n_1 = 5$

$n_2 =4$

$\therefore$

$\dfrac{1}{\lambda_A} = R\bigg(\dfrac{1}{4^2}-\dfrac{1}{5^2}\bigg)$

$\implies \ \lambda_A = \dfrac{44.44}{R}$
(B) :

$n_1 = 4$

$n_2 =3$

$\therefore$

$\dfrac{1}{\lambda_B} = R\bigg(\dfrac{1}{3^2}-\dfrac{1}{4^2}\bigg)$

$\implies \ \lambda_B = \dfrac{20.57}{R}$
(C) :

$n_1 = 3$

$n_2 =2$

$\therefore$

$\dfrac{1}{\lambda_C} = R\bigg(\dfrac{1}{2^2}-\dfrac{1}{3^2}\bigg)$

$\implies \ \lambda_C = \dfrac{7.2}{R}$
(D) :

$n_1 = 2$

$n_2 =1$

$\therefore$

$\dfrac{1}{\lambda_D }= R\bigg(\dfrac{1}{1^2}-\dfrac{1}{2^2}\bigg)$

$\implies \ \lambda_D = \dfrac{1.33}{R}$
Thus wavelength is minimum in

$n=2$ to

$n=1$ .

Total energy of electron in an excited state of hydrogen atom is

$-3.4 eV$ . The kinetic and potential energy of electron in this state

0%
$K=-3.4 eV$ $U=-6.8 eV$
0%
$K=3.4 eV$ $U=-6.8 eV$
0%
$K=-6.8 eV$ $U=+3.4 eV$
0%
$K=+10.2 eV$ $U=-13.6 eV$

Explanation

Total energy of electron,

$T = -3.4eV$ (Given)

Kinetic energy of electron,

$K = |T| =3.4eV$

Potential energy of electron,

$U = 2T = -6.8eV$

An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision

0%
must be elastic
0%
may be partially elastic
0%
must be completely inelastic
0%
may be completely inelastic.

Explanation

In hydrogen atom, $E_n = \dfrac{-13.6}{n^2} eV$

$n = 1; E_1 = -13.6 eV$

$n = 2; E_2 = \dfrac{-13.6}{4} eV$

First exited state = $3.4 eV$

$E_2 – E_1 = 10.2 eV$

Since energy difference is 10.2eV which is greater than K.E. 5 eV, the electron exites the hydrogen atom, hence the collision must be elastic as there is no loss in energy of the electron.

When a photon stimulates the emission of another photon, the two photons have

0%
same energy
0%
same direction
0%
same phase
0%
same wavelength.

Explanation

When a photon stimulates another photon then this process is called stimulated emission. It is the process used in

$LASER$

Then the energy of emitted photon will be equal to energy of incident photon that is why we say it Amplification because as number of photons with same energy get increased.

As energy is same so wavelength will also be same.

Same direction because in LASER we get

$unidirectional$ beam.

Same phase because in LASER we get

$coherence$ beam.

$E_P$ and

$E_k$ represent potential energy and kinetic energy respectively, of an orbital electron, then according to Bohr's theory:

0%
$E_k= -E_p/2$
0%
$E_k= -E_p$
0%
$E_k= -2E_p$
0%
$E_k= 2E_p$

Explanation

Potential energy of an orbital electron

$E_p=-\dfrac{Ze^2}{r}$

Kinetic energy of an electron

$E_k=\dfrac{Ze^2}{2r}$ , where Z=atomic number

So,

$E_k=-\dfrac{E_p}{2}$

Which of the following spectral series of hydrogen atom is lying in visible range of electromagnetic wave?

0%
Paschen series
0%
Pfund series
0%
Lyman series
0%
Balmer series

When electron jumps from

$n=4$ level to

$n=1$ level, the angular momentum of electron changes by

0%
$\dfrac{h}{2\pi}$
0%
$\dfrac{2h}{2\pi}$
0%
$\dfrac{3h}{2\pi}$
0%
$\dfrac{4h}{2\pi}$

Explanation

Angular momentum of electron in nth state,

$J_n = \dfrac{nh}{2\pi}$

Change in angular momentum,

$\Delta J = (n_1-n_2)\dfrac{h}{2\pi}$

$\therefore$

$\Delta J = \dfrac{(4-1)h}{2\pi} = \dfrac{3h}{2\pi}$

White light is passed through sodium vapours contained in a thin walled glass flask and the transmitted light is examined with the help of a spectrometer. The spectrum so obtained is.

0%
Absorption spectrum
0%
Solar spectrum
0%
Band spectrum
0%
Continuous spectrum

If the minimum energy of photons needed to produce photoelectric effect is 3eV, bombarding the photoelectric material by a number of photons of 2.5eV also one can get

0%
photoelectrons of the same kinetic energy
0%
photoelectrons of the higher kinetic energy
0%
higher current
0%
none

Explanation

Work function of metal is given as

$\phi = 3 \ eV$
Energy of incident photon

$E_{incident} = 2.5 \ eV$
Emission of photo electrons from the metal surface takes place only when the energy of the incident photon is greater than the work function of the metal surface.
Since here, energy of incident photon is less than work function of metal surface. So emission of electron is not possible.

When a beam of white light is passed through sodium vapours and then through a spectrometer, spectrum so obtained has two dark lines present in the yellow region. This spectrum is called

0%
band spectrum
0%
continuous spectrum
0%
absorption spectrum of sodium
0%
emission spectrum of sodium

In the hydrogen atom an electron is moving in nth orbit. The circumference s of the orbit and the de Broglie wavelength, of the moving electron are related by the equation

0%
$S=n \lambda$
0%
$S=\dfrac{\lambda}{n}$
0%
$S=\dfrac{n}{\lambda}$
0%
None of these

Explanation

It is a postulate by Bohr , written as

$2 \pi r =n \lambda$

circumference

$S=n\times wavelength$

When electron in hydrogen atom jumps from second orbit to first orbit, the wavelength of radiation emitted is

$\lambda$ . When electron jumps from third orbit to first orbit, the wavelength of emitted radiation would be

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$\cfrac { 27 }{ 32 } \lambda$
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$\cfrac { 32 }{ 27 } \lambda$
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$\cfrac { 2 }{ 3 } \lambda$
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$\cfrac { 3 }{ 2 } \lambda$

Explanation

Wavelength of photon emitted

$\dfrac{1}{\lambda} = R \bigg(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\bigg)$

$(\because Z = 1)$

Transaction from second orbit to first orbit :

We have,

$n_1 = 2$

$n_2 = 1$

$\therefore$

$\dfrac{1}{\lambda} = R \bigg(\dfrac{1}{1^2} - \dfrac{1}{2^2}\bigg)$

$\implies R = \dfrac{4}{3\lambda}$

Transaction from third orbit to first orbit :

We have,

$n_1 = 3$

$n_2 = 1$

$\therefore$

$\dfrac{1}{\lambda'} = R \bigg(\dfrac{1}{1^2} - \dfrac{1}{3^2}\bigg)$

$\implies \lambda' =\dfrac{9}{8R}= \dfrac{9}{8}\times \dfrac{3\lambda}{4} = \dfrac{27}{32}\lambda$

Radius of the second Bohr orbit of a singly ionised helium atom is

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$0.53 A^o$
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$1.06 A^o$
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$0.265 A^o$
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$0.132 A^o$

Explanation

Radius of nth Bohr orbit

$r_n = \dfrac{0.529 n^2}{Z} \ A^o$
For ionised Helium atom

$Z = 2$
Second Bohr orbit

$n=2$

$\implies \ r_2 = \dfrac{0.529\times 2^2}{2} = 1.06 \ A^o$

The wave length of K-ray line of an anti cathode element of atomic number Z is nearly proportional to:

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$Z^2$
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$(Z-1)^2$
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1/(Z-1)
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1/ $(Z-1)^2$

Explanation

Frequency of K-ray

$\nu\propto (Z-1)^2$
We know that

$\lambda = \dfrac{v}{\nu}$
Thus

$\lambda \propto \dfrac{1}{(Z-1)^2}$

If the electron in the hydrogen atoms is excited to n = 5 state, the number of frequencies present in the radiation emitted is :

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4
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5
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8
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10

Explanation

The given excited state

$n = 5$
Thus number of frequencies (or spectral lines)

$N = \dfrac{n(n-1)}{2}$

$\implies \ N = \dfrac{5\times (5-1)}{2} =10$

CBSE Questions for Class 12 Medical Physics Atoms Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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